STRUCTURE OF CHROMIUM ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy). (August 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks discovered by Gell-Mann and Zweig I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Naturally occurring chromium (Cr) is composed of four stable isotopes; Cr-50, Cr-52, Cr-53, and Cr-54 with Cr-52 being the most abundant (83.789% natural abundance). Cr-50 is suspected of decaying by β+β+ to Ti-50 with a half-life of (more than) 1.8x1017 years. Twenty-two radioisotopes, all which are entirely synthetic, have been characterized with the most stable being Cr-51 with a half-life of 27.7 days. All of the remaining radioactive isotopes have half-lives that are less than 24 hours and the majority of these have half-lives that are less than 1 minute, the least stable being Cr-66 with a half-life of 10 milliseconds. This element also has 2 meta states. WHY Cr-50, Cr-52, Cr-53 and Cr-54 ARE STABLE NUCLIDES After a careful analysis of the structure of atomic nuclei I discovered that the beta decay is due to the fact that in unstable nuclei there exist single horizontal pn bonds of weak binding energy leading to the beta decay. For example in my paper STRUCTURE AND BINDING OF H3 AND He3 using the diagram of the structure of the H3 one sees that it is unstable because the two neutrons make single np bonds, while the He3 is stable because the one neutron between the two protons makes two np bonds per neutron. On the other hand the pp repulsions of long range lead to the instability when we have a small number of pn bonds per nucleon. For comparing the structure between the stable Cr-52 and the unstable Cr-48 you may read my STRUCTURE OF Cr-52 AND Cr-48 . A careful analysis of such a comparison shows that the stability of Cr-50 , Cr-52 and Cr-54 with S=0 (even number of neutrons) is due to the high symmetry of the structure having enough number of extra neutrons with two bonds per neutron ( one horizontal and one strong vertical bond) able to overcome the pp and nn repulsions. Whereas the unstable Cr-48 with 24 protons and 24 neutrons has not any extra neutron for making strong extra bonds for increasing the binding energies of bonds. Also in the following diagram of the unstable Cr-49 with S = -5/2 (odd number of neutrons) you can see that it has only the one extra n25(-1/2) which makes two weak horizontal bonds unable to give extra binding energy to bonds for overcoming the repulsions of pp and nn systems. However the Cr-53 with S = -3/2 based on the structure of Cr-49 with S = -5/2 is a stable nuclide because it has two more extra neutrons of positive spins which make two bonds per neutron ( one weak and one strong vertical bond) able to give enough energy to bonds for overcoming the repulsions . Under this condition the total spin of Cr-53 is given by S = -5/2 +2(+1/2) = -3/2. ' DIAGRAM OF THE UNSTABLE Cr-49 WITH S = -5/2' This structure has the core of Mg-24 with S=0 having 6 horizontal planes of opposite spins like the +HP1, -HP2, +HP3 -HP4, +HP5 and -HP6. Here the deuterons of the alpha particles like the p19 n19 and n20p20 are not shown because they are behind the n6p6 and p8n8 respectively. Also the n17p17and p18n18 are not shown here because they are in front of p5n5 and n7p7 respectively. Moreover the extra n25(-1/2 ) which makes two weak horizontal bonds between the protons of alpha particles is not shown because it is behind the p16. Since the p24n24 and n23p23 of -HP6 give S =-2 we get a total S = -5/2. ' n24……….p12.........n12..........p23' ' -HP6 p24.........n11..........p11.........n23 ' ' p22.........n10..........p10...........n21' ' +HP5 n22..........p9............n9..........p21 ' '' '' n14..........p8.......... n8............p16 '' ''-HP4'' p14..........n7............p7...........n16 ' ''' p13..........n6...........p6............n15 ' +HP3 n13..........p5...........n5...........p15 ' '' '' p4............n4 ' -HP2 n3...........p3 ' ' n2............p2' ' +HP1 p1...........n1 '' '' ' NUCLEAR STRUCTURE OF THE UNSTABLE Cr-51 WITH S =-7/2 ''' After a careful analysis we found that the unstable structure of the Cr-51 with S=-7/2 is based on the structure of Cr-49 with S=-5/2. Here the Cr-51 has two more extra neutrons of negative spins which make two weak horizontal bonds unable to increase enough the energies of bonds. Under this condition the total spins is given by S = -5/2 +2(-1/2) = -7/2 '''NUCLEAR STRUCTURE OF THE UNSTABLE Cr-55, Cr-57, Cr-59, Cr-61, Cr-63, Cr-65 AND Cr-67 After a detailed analysis we found that the structure of the above unstable nuclides is based on the structure of Cr-53 with S =-3/2. For example the Cr-67 with S=-1/2 has two extra nucleons with single bonds of positive spins and 12 extra neutrons of opposite spins giving S=0. Thus the total spin is given by S = -3/2 +2(+1/2) + 0 = -1/2 NUCLEAR STRUCTURE OF THE UNSTABLE Gr-47 AND Cr-45 In the absence of neutrons we see that the structure of the above unstable nuclides is based on the structure of Gr-49 with S=-5/2. For example in the structure of the Cr-45 with S =-7/2 we see that there are two absent neutrons of positive spins and two absent neutrons of opposite spins giving S=0. That is S = -5/2 - 2(+1/2) - 0 = -7/2. NUCLEAR STRUCTURE OF THE UNSTABLE Cr-43 WITH S = +3/2 In the absence of 6 neutrons we see that the structure of Cr-43 is based on a structure which is similar to the structure of Cr-49. In this case all nucleons of Cr-49 change the spins. For example we have -HP1, +HP2, -HP3, +HP4, -HP5 and +HP6 giving S = +5/2. Under this change of spins in the structure of Cr-43 with S = +3/2 we have two absent neutrons of positive spins and 4 absent neutrons of opposite spins giving S=0. That is S = +5/2 - 2 (+1/2) - 0 = +3/2. ' ' NUCLEAR STRUCTURE OF Cr-54, Cr-56, Cr-58, Cr-60, Cr-62, Cr-64 and Cr-66 WITH S=0 The structure of the above unstable nuclides is based on the structure of the stable Cr-54 with S=0. For example the Cr-66 with S=0 has 12 more extra neutrons of single bonds with opposite spins giving S=0. NUCLEAR STRUCTURE OF THE UNSTABLE Cr-46 AND Cr-44 WITH S=0 In the absence of neutrons the structure of the above nuclides is based on the structure of Cr-48 with S=0. For example in the structure of Cr-44 with S=0 we have 4 absent neutrons of opposite spins giving S=0. Category:Fundamental physics concepts